Do do you this one please
$\displaystyle \int \frac {1 + x}{1 + x^2}~dx = \int \frac {1}{1 + x^2}~dx + \int \frac {x}{1 + x^2}~dx$
$\displaystyle = \arctan x + \int \frac {x}{1 + x^2}~dx$
We proceed by substitution for the second integral:
Let $\displaystyle u = 1 + x^2$
$\displaystyle \Rightarrow du = 2x ~dx$
$\displaystyle \Rightarrow \frac {1}{2} du = x ~dx$
$\displaystyle \Rightarrow \int \frac {1 + x}{1 + x^2}~dx = \arctan x + \frac {1}{2} \int \frac {1}{u} ~du$
$\displaystyle = \arctan x + \frac {1}{2} \ln u + C$
$\displaystyle = \boxed { \arctan x + \frac {1}{2} \ln \left( 1 + x^2 \right) + C }$
EDIT: You're too fast for me red_dog!
I will do the indefinite integral for the second, i leave plugging in the limits to you
$\displaystyle \int \tan^3 x ~dx = \int \tan^2 x \tan x ~dx$
$\displaystyle = \int \left( \sec^2 x - 1 \right) \tan x ~dx$
$\displaystyle = \int \left( \sec^2 x \tan x - \tan x \right)~dx$
$\displaystyle = \int \sec^2 x \tan x ~dx - \int \tan x ~dx$
$\displaystyle = \int \sec^2 x \tan x ~dx - \ln | \sec x | + C$
We proceed by substitution for the first integral:
Let $\displaystyle u = \tan x$
$\displaystyle \Rightarrow du = \sec^2 x ~dx$
$\displaystyle \Rightarrow \int \tan^3 x ~dx = \int u ~du - \ln | \sec x | + C$
$\displaystyle = \frac {1}{2} u^2 - \ln | \sec x | + C$
$\displaystyle = \boxed { \frac {1}{2} \tan^2 x - \ln | \sec x | + C }$
Ah, i never noticed that (never paid attention to the limits, i planned on finding the indefinate integral from the start). That's true, the integral is 0. However, if what you actually wanted was the area, you would find $\displaystyle 2 \int_{0}^{ \frac { \pi}{6}} \tan^3 x~dx$
Another possible solution, similar to Jhevon's is:
$\displaystyle \int{tan^{3}(x)}dx=\int{(sec^{2}(x)-1)tan(x)}dx=sec^{2}(x)tan(x)-tan(x)$
$\displaystyle \int{sec^{2}(x)tan(x)}dx$
Now, let $\displaystyle u=sec(x), \;\ du=sec(x)tan(x)dx$
$\displaystyle \int{u}du=\frac{1}{2}u^{2}=\frac{1}{2}sec^{2}(x)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\displaystyle \int{tan(x)}dx$
$\displaystyle \int\frac{sin(x)}{cos(x)}dx$
Let $\displaystyle u=cos(x), \;\ -du=sin(x)dx$
$\displaystyle -\int\frac{1}{u}du=ln(u)=-ln(cos(x))$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So, we have $\displaystyle \boxed{\frac{1}{2}sec^{2}(x)+ln(cos(x))+C}$
$\displaystyle \frac{1}{2}tan^{2}(x)-ln(sec(x))+\frac{1}{2}=\frac{1}{2}sec^{2}(x)+ln(co s(x))$